We just introduced the concept of the Frechet-derivative in class and I'm trying to grasp how to do it so I want to know wether my understanding is correct.
Let's consider $$f:\mathbb{R}^{N\times N} \to \mathbb{R}^{N\times N}, X\mapsto X^2$$
The continous, linear function $T$ is the Frechet derivative of $f$, if $$f(X+ \epsilon) = f(X) + T\epsilon + r(\epsilon)$$ and $$\lim_{||\epsilon||\to0}\frac{r(\epsilon)}{||\epsilon||} = 0$$
If I now want to calculate this for $f$, I get $$f(X + \epsilon) = (X + \epsilon)^2 = X^2 + X\epsilon + \epsilon X + \epsilon^2$$
$\epsilon^2$ satisfies the limit condition, $X^2$ is $f(x)$, so I think that $X\epsilon + \epsilon X$ = Tx
But how can I explictly find this $T$?
$T$ is exactly the linear map $\mathbb R^{N\times N} \to \mathbb R^{N\times N}$ given by
$$T\epsilon = X\epsilon + \epsilon X, \ \ \ \forall \epsilon \in \mathbb R^{N\times N}.$$
You cannot be more explicit then that, since $X\in \mathbb R^{N\times N}$ is arbitrary. This is the Frechet-derivative at $X$.